Help Understanding Proof of Eisenstein's Criterion
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The proof begins like this.
So we suppose $f(x)= a_0+a_1x+...+a_nx^n$ is reducible and thus
$f(x) = g(x)h(x) = (b_0+b_1x+...+b_rx^r)(c_0+c_1x+...+c_sx^s$).
Then $a_0=b_0c_0$. By the definition of Eisenstein's criterion $pmid a_0$ and so $pmid b_0$ or $pmid c_0$. Lets say $pmid b_0$. "Since $p^2$ does not divide $a_0$ (by definition of Eisentein's criterion), we see that $c_0$ is not divisible by $p$."
I don't understand that last statement. How does $c_0$ not being divisible by $p$ follow from the fact that $p^2$ does not divide $a$?
elementary-number-theory polynomials
edited 2 days ago
Carmeister
2,5492920
asked 2 days ago
s_healy
103
add a comment |
up vote
0
down vote
favorite
The proof begins like this.
So we suppose $f(x)= a_0+a_1x+...+a_nx^n$ is reducible and thus
$f(x) = g(x)h(x) = (b_0+b_1x+...+b_rx^r)(c_0+c_1x+...+c_sx^s$).
Then $a_0=b_0c_0$. By the definition of Eisenstein's criterion $pmid a_0$ and so $pmid b_0$ or $pmid c_0$. Lets say $pmid b_0$. "Since $p^2$ does not divide $a_0$ (by definition of Eisentein's criterion), we see that $c_0$ is not divisible by $p$."
I don't understand that last statement. How does $c_0$ not being divisible by $p$ follow from the fact that $p^2$ does not divide $a$?
elementary-number-theory polynomials
edited 2 days ago
Carmeister
2,5492920
asked 2 days ago
s_healy
103
2
If $p| c_0$ and $p|b_0,$ then $p^2(c_0/p)(b_0/p)=a_0$, hence $p^2| a_0$. By contrapositive if $p|b_0$, then $pnot|c_0.$
– Melody
2 days ago
thank you so much!
– s_healy
2 days ago
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
The proof begins like this.
So we suppose $f(x)= a_0+a_1x+...+a_nx^n$ is reducible and thus
$f(x) = g(x)h(x) = (b_0+b_1x+...+b_rx^r)(c_0+c_1x+...+c_sx^s$).
Then $a_0=b_0c_0$. By the definition of Eisenstein's criterion $pmid a_0$ and so $pmid b_0$ or $pmid c_0$. Lets say $pmid b_0$. "Since $p^2$ does not divide $a_0$ (by definition of Eisentein's criterion), we see that $c_0$ is not divisible by $p$."
I don't understand that last statement. How does $c_0$ not being divisible by $p$ follow from the fact that $p^2$ does not divide $a$?
elementary-number-theory polynomials
edited 2 days ago
Carmeister
2,5492920
asked 2 days ago
s_healy
103
The proof begins like this.
So we suppose $f(x)= a_0+a_1x+...+a_nx^n$ is reducible and thus
$f(x) = g(x)h(x) = (b_0+b_1x+...+b_rx^r)(c_0+c_1x+...+c_sx^s$).
Then $a_0=b_0c_0$. By the definition of Eisenstein's criterion $pmid a_0$ and so $pmid b_0$ or $pmid c_0$. Lets say $pmid b_0$. "Since $p^2$ does not divide $a_0$ (by definition of Eisentein's criterion), we see that $c_0$ is not divisible by $p$."
I don't understand that last statement. How does $c_0$ not being divisible by $p$ follow from the fact that $p^2$ does not divide $a$?
elementary-number-theory polynomials
elementary-number-theory polynomials
edited 2 days ago
Carmeister
2,5492920
asked 2 days ago
s_healy
103
edited 2 days ago
Carmeister
2,5492920
asked 2 days ago
s_healy
103
edited 2 days ago
Carmeister
2,5492920
edited 2 days ago
Carmeister
2,5492920
edited 2 days ago
Carmeister
2,5492920
2,5492920
asked 2 days ago
s_healy
103
asked 2 days ago
s_healy
103
asked 2 days ago
s_healy
103
103
2
If $p| c_0$ and $p|b_0,$ then $p^2(c_0/p)(b_0/p)=a_0$, hence $p^2| a_0$. By contrapositive if $p|b_0$, then $pnot|c_0.$
– Melody
2 days ago
thank you so much!
– s_healy
2 days ago
add a comment |
2
If $p| c_0$ and $p|b_0,$ then $p^2(c_0/p)(b_0/p)=a_0$, hence $p^2| a_0$. By contrapositive if $p|b_0$, then $pnot|c_0.$
– Melody
2 days ago
thank you so much!
– s_healy
2 days ago
2
2
If $p| c_0$ and $p|b_0,$ then $p^2(c_0/p)(b_0/p)=a_0$, hence $p^2| a_0$. By contrapositive if $p|b_0$, then $pnot|c_0.$
– Melody
2 days ago
If $p| c_0$ and $p|b_0,$ then $p^2(c_0/p)(b_0/p)=a_0$, hence $p^2| a_0$. By contrapositive if $p|b_0$, then $pnot|c_0.$
– Melody
2 days ago
thank you so much!
– s_healy
2 days ago
thank you so much!
– s_healy
2 days ago
add a comment |
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2
If $p| c_0$ and $p|b_0,$ then $p^2(c_0/p)(b_0/p)=a_0$, hence $p^2| a_0$. By contrapositive if $p|b_0$, then $pnot|c_0.$
– Melody
2 days ago
thank you so much!
– s_healy
2 days ago